Question 157467
Let {{{x}}}= 1st no.
{{{y}}} = 2nd no.
1st condition: {{{x+y=10}}} ------------------------> eqn 1
2nd condition: {{{3x=4y+2}}} -----------------------> eqn 2
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In 2nd condition, 3 times one of them is more than 4 times the other by 2, right?
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In eqn 1 we get, {{{x=10-y}}} ----------------------> eqn 3, substitute in eqn 2
{{{3(10-y)=4y+2}}}
{{{30-3y=4y+2}}}
{{{30-2=4y+3y}}}
{{{28=7y}}} ----> {{{cross(28)4/cross(7)=cross(7)y/cross(7)}}}
{{{y=4}}} ------------> 2nd no.
Go back eqn 3: {{{x=10-4=6}}} -------------> 1st no.
It satisfy the 1st condition:{{{6+4=10}}}
Thank you,
Jojo